Answer to Question #4189 in Calculus for shane duque

Question

A man 6 ft. tall walks away from a lamppost 16 ft. high at a rate of 5 mi per hr. How fast does the of his shadow move?

Answer 1

Suppose a man was in point A and the shadow was in point S₁ at time t = t₀. After some time t, the man came to the point B and the shadow – to the point S₂. OL is a lamppost.
As we want to evaluate the speed of the shadow, we need to find:
(S₁ S₀)/t = (S₁ B+BS₂)/t


Angles <OS₁L and <CS₁A are equal, so:
OL/(LA+AS₁ )=AC/(AS₁ )
AS₁*H = AS₁*h + LA*h
Where H is the height of the lamppost (H = 16ft) and h is the man’s height (h = 6ft).
AS₁ = LA*h/(H-h)
So, S₁ B = AB - AS₁ = Vt - LA*h/(H-h)
Angles <OS₂ L and <DS₂ B are equal, so:
OL/(LA+AB+BS₂ ) = BD/(BS₂ )
BS₂*H = LA*h + Vt*h + BS₂*h
BS₂ = h(LA+Vt)/(H-h)
Then:
(S₁ S₂)/t = (S₁ B+BS₂)/t = V - LA/t*h/(H - h) + h(LA + Vt)/(H - h)t = V + (h*LA + Vth - LA*h)/((H - h)t)
= V + V*h/(H - h) = VH/(H - h)
The speed of the shadow is 5*16/(16-6)=8 miles/hour